Express y=3x^2-9x +1 in the form y=a(x+b)^2+c so i completed the square, but i got y=3(x-3/2)^2-77/4 .... its the wrong answer

Question

Express y=3x^2-9x +1 in the form y=a(x+b)^2+c

so i completed the square, but i got y=3(x-3/2)^2-77/4 .... its the wrong answer ><

anonymous · Answer

@ganeshie8 d'you know anything about completing squares for graphs? :3

ganeshie8 · Answer

$\large   y=3x^2-9x +1 $

$\large   y=3(x^2-3x) +1 $

$\large   y=3(x^2-3x + \frac{9}{4}) +1 - 3\times \frac{9}{4} $

$\large   y=3(x - \frac{3}{2})^2 +1 - \frac{27}{4} $

$\large   y=3(x - \frac{3}{2})^2 - \frac{23}{4} $

ganeshie8 · Answer

your last term is wrong that all.. go through above to see where exactly you did the mistake

anonymous · Answer

hang on... why 3 x 9/4? o-o

isn't it 1-(-9/2)^2?

ganeshie8 · Answer

inside the parenthesis, you're adding `9/4` right ?

anonymous · Answer

ohh shoot the square method can only be done if the coefficient of x^2 is one, right?

ganeshie8 · Answer

it still works here also 
notice that 9/4 = (3/2)^2

anonymous · Answer

yes, but i've never done a rule that needed me to multiply it by the x coefficient ><

ganeshie8 · Answer

inside the parenthesis you're adding (3/2)^2
so you need to subtract that outside the parenthesis :  -(3/2)^2
however, since the whole parenthesis was multiplied by 3, you need to subtract  :  -3 (3/2)^2 = -3*9/4

anonymous · Answer

i only factorised 3 from the 3(x-3/2)^2, so shouldn't i only add it back to that one?

ganeshie8 · Answer

one way to make sense of above is to multiply again, simplify and see if you get back to the earlier equation

anonymous · Answer

okay..

so when i complete the square for an equation with the x^2 coefficient more than 1, i have to multiply the back number with the coefficient?

anonymous · Answer

what just happened o-o

sorry i forgot to reply i fell asleep /shot

anonymous · Answer

wait and why the talk about code o-O

ganeshie8 · Answer

your thread was hijacked by @dan815 earlier

ganeshie8 · Answer

suppose you have :
$\large   a(x+b)$

ganeshie8 · Answer

if you add 2 and subtract inside the parenthesis, nothing changes :

$\large a(x+b) =  a(x+b\color{red}{+2-2})$

right ?

ganeshie8 · Answer

by distributing, you can kick that -2 out of the parenthesis :

$\large a(x+b) =  a(x+b\color{red}{+2}) \color{Red}{-} a(\color{red}{2})$